Question

show that:
(1+cotx)/cscx - secx/(tanx + cotx) = cosx

Answer 1

if you want a hint, i would rewrite everything without sines and cosines to make the algebra easier
put
\[\cos(x)=a,\sin(x)=b\]

Answer 2

then you have
\[\frac{1+\frac{a}{b}}{\frac{1}{b}}-\frac{\frac{1}{a}}{\frac{b}{a}+\frac{a}{b}}=a\]

Answer 3

Input: 1 + cos(x) / csc(x) - sec(x) / tan(x) + cot(x) = cos(x)
Results: True.

Answer 4

do a bunch of algebra unencumbered by sines and cosines
at some point you will probably need that
\[\cos^2(x)+\sin^2(x)=1\] so if you see
\[a^2+b^2\] replace it by 1

this is mostly just annoying algebra, there is little or no trig in it

Answer 5

i would show the solution..the equation thingie is freaking weird..

Answer 6

i can write it out for you if you like, that is i can write out the algebra

Answer 7

...ok..............

Answer 8

\[\frac{1+\frac{a}{b}}{\frac{1}{b}}-\frac{\frac{1}{a}}{\frac{b}{a}+\frac{a}{b}}=a\] multiply top and bottom of first fraction by b and second one by ab get
\[b+a-\frac{b}{b^2+a^2}=a\] now replace
\[a^2+b^2=1\] get
\[b+a-b=a\] and you are done

Answer 9

if you have to turn this in rewrite exactly what i wrote replacing a by cosine and b by sine

Answer 10

(1+cotx)/cscx - secx/(tanx + cotx)
= 1/cscx + cotx/cscx - secx/ {[(sinx)^2 + (cosx)^2 ] / cosxsinx}
= sinx + cotxsecx - secx / (1/sinxcosx)
= sinx + (cosx/sinx)/(1/sinx) - secx(sinxcosx)
=sinx + cosx - sinx
= cosx